\(\int \sec ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) [236]

Optimal result

Integrand size = 25, antiderivative size = 76 \[ \int \sec ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 \sqrt {b} f}+\frac {\tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 f} \] Output:

1/2*(a+b)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/b^(1/2)/f 
+1/2*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/f
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(210\) vs. \(2(76)=152\).

Time = 1.22 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.76 \[ \int \sec ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)} \left (\sqrt {2} (a+b) \text {arctanh}\left (\frac {\sqrt {\frac {b \sin ^2(e+f x)}{a+b}}}{\sqrt {\frac {a+b-a \sin ^2(e+f x)}{a+b}}}\right ) \cos ^2(e+f x) \sqrt {\frac {a+2 b+a \cos (2 (e+f x))}{a+b}}+(a+2 b+a \cos (2 (e+f x))) \sqrt {\frac {b \sin ^2(e+f x)}{a+b}}\right ) \tan (e+f x)}{\sqrt {2} f (a+2 b+a \cos (2 (e+f x)))^{3/2} \sqrt {\frac {b \sin ^2(e+f x)}{a+b}}} \] Input:

Integrate[Sec[e + f*x]^2*Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

(Sqrt[a + b*Sec[e + f*x]^2]*Sqrt[a + b - a*Sin[e + f*x]^2]*(Sqrt[2]*(a + b 
)*ArcTanh[Sqrt[(b*Sin[e + f*x]^2)/(a + b)]/Sqrt[(a + b - a*Sin[e + f*x]^2) 
/(a + b)]]*Cos[e + f*x]^2*Sqrt[(a + 2*b + a*Cos[2*(e + f*x)])/(a + b)] + ( 
a + 2*b + a*Cos[2*(e + f*x)])*Sqrt[(b*Sin[e + f*x]^2)/(a + b)])*Tan[e + f* 
x])/(Sqrt[2]*f*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2)*Sqrt[(b*Sin[e + f*x]^2 
)/(a + b)])
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4634, 211, 224, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sec ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (e+f x)^2 \sqrt {a+b \sec (e+f x)^2}dx\)

\(\Big \downarrow \) 4634

\(\displaystyle \frac {\int \sqrt {b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {1}{2} (a+b) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{2} (a+b) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 \sqrt {b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

Input:

Int[Sec[e + f*x]^2*Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

(((a + b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/ 
(2*Sqrt[b]) + (Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/2)/f
 

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4634

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) 
)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f 
Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), 
x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ 
[m/2] && IntegerQ[n/2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(508\) vs. \(2(64)=128\).

Time = 23.77 (sec) , antiderivative size = 509, normalized size of antiderivative = 6.70

method	result	size
default	\(\frac {\sqrt {a +b \sec \left (f x +e \right )^{2}}\, \left (\ln \left (\frac {4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a -4 a -4 b}{\sin \left (f x +e \right )+1}\right ) b^{\frac {3}{2}} \cos \left (f x +e \right )+\ln \left (\frac {4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a -4 a -4 b}{\sin \left (f x +e \right )+1}\right ) \sqrt {b}\, \cos \left (f x +e \right ) a +\ln \left (-\frac {4 \left (\sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+\sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-\sin \left (f x +e \right ) a +a +b \right )}{\sin \left (f x +e \right )-1}\right ) b^{\frac {3}{2}} \cos \left (f x +e \right )+\ln \left (-\frac {4 \left (\sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+\sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-\sin \left (f x +e \right ) a +a +b \right )}{\sin \left (f x +e \right )-1}\right ) \sqrt {b}\, \cos \left (f x +e \right ) a +\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b \left (2 \sin \left (f x +e \right )+2 \tan \left (f x +e \right )\right )\right )}{4 f b \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\)	\(509\)

Input:

int(sec(f*x+e)^2*(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/4/f/b*(a+b*sec(f*x+e)^2)^(1/2)/(1+cos(f*x+e))/((b+a*cos(f*x+e)^2)/(1+cos 
(f*x+e))^2)^(1/2)*(ln(4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/ 
2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+ 
e)*a-a-b)/(sin(f*x+e)+1))*b^(3/2)*cos(f*x+e)+ln(4*(b^(1/2)*((b+a*cos(f*x+e 
)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos 
(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*b^(1/2)*cos(f*x+e)*a+l 
n(-4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+b^(1/ 
2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+ 
e)-1))*b^(3/2)*cos(f*x+e)+ln(-4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e) 
)^2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)- 
sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*b^(1/2)*cos(f*x+e)*a+((b+a*cos(f*x+e)^2) 
/(1+cos(f*x+e))^2)^(1/2)*b*(2*sin(f*x+e)+2*tan(f*x+e)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (64) = 128\).

Time = 0.14 (sec) , antiderivative size = 320, normalized size of antiderivative = 4.21 \[ \int \sec ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\left [\frac {{\left (a + b\right )} \sqrt {b} \cos \left (f x + e\right ) \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) + 4 \, b \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{8 \, b f \cos \left (f x + e\right )}, \frac {{\left (a + b\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right ) + 2 \, b \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{4 \, b f \cos \left (f x + e\right )}\right ] \] Input:

integrate(sec(f*x+e)^2*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[1/8*((a + b)*sqrt(b)*cos(f*x + e)*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 
 + 8*(a*b - b^2)*cos(f*x + e)^2 + 4*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x 
+ e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8 
*b^2)/cos(f*x + e)^4) + 4*b*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*si 
n(f*x + e))/(b*f*cos(f*x + e)), 1/4*((a + b)*sqrt(-b)*arctan(-1/2*((a - b) 
*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/c 
os(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e)))*cos(f*x + e) + 2 
*b*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(b*f*cos(f*x 
+ e))]
 

Sympy [F]

\[ \int \sec ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \sqrt {a + b \sec ^{2}{\left (e + f x \right )}} \sec ^{2}{\left (e + f x \right )}\, dx \] Input:

integrate(sec(f*x+e)**2*(a+b*sec(f*x+e)**2)**(1/2),x)
 

Output:

Integral(sqrt(a + b*sec(e + f*x)**2)*sec(e + f*x)**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.91 \[ \int \sec ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\frac {a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + \sqrt {b \tan \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right )}{2 \, f} \] Input:

integrate(sec(f*x+e)^2*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

1/2*(a*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/sqrt(b) + sqrt(b)*arcsinh(b 
*tan(f*x + e)/sqrt((a + b)*b)) + sqrt(b*tan(f*x + e)^2 + a + b)*tan(f*x + 
e))/f
 

Giac [F]

\[ \int \sec ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{2} \,d x } \] Input:

integrate(sec(f*x+e)^2*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(b*sec(f*x + e)^2 + a)*sec(f*x + e)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \frac {\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{{\cos \left (e+f\,x\right )}^2} \,d x \] Input:

int((a + b/cos(e + f*x)^2)^(1/2)/cos(e + f*x)^2,x)
 

Output:

int((a + b/cos(e + f*x)^2)^(1/2)/cos(e + f*x)^2, x)
 

Reduce [F]

\[ \int \sec ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2}d x \] Input:

int(sec(f*x+e)^2*(a+b*sec(f*x+e)^2)^(1/2),x)
 

Output:

int(sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**2,x)